(x³+x²y+xy²+y³)(x-y)

=x(x³+x²y+xy²+y³)-y(x³+x²y+xy²+y³)

=x⁴+x³y+x²y²+xy³-x³y-x²y²+xy³+y⁴

= x⁴+y⁴

đề sai bạn xem lại đề

Ta có: \(\left(x^3-x^2y+xy^2-y^3\right)\left(x+y\right)\)

\(=\left[x^2\left(x-y\right)+y^2\left(x-y\right)\right]\left(x+y\right)\)

\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)

\(=x^4-y^4=2^4-\left(\dfrac{1}{2}\right)^4=16-\dfrac{1}{16}=\dfrac{255}{16}\)

Với x ≥ 0; y ≥ 0 thì x + y ≥ 0

Ta có: x3 + y3 ≥ x2y + xy2

⇔ (x3 + y3) – (x2y + xy2) ≥ 0

⇔ (x + y)(x2 – xy + y2) – xy(x + y) ≥ 0

⇔ (x + y)(x2 – xy + y2 – xy) ≥ 0

⇔ (x + y)(x2 – 2xy + y2) ≥ 0

⇔ (x + y)(x – y)2 ≥ 0 (Luôn đúng vì x + y ≥ 0 ; (x – y)2 ≥ 0)

Dấu « = » xảy ra khi (x – y)2 = 0 ⇔ x = y.

Sửa đề: \(A=x^3+x^2y-xy^2-y^3+x^2-y^2+2x+2y+3\)

\(A=x^2\left(x+y\right)-y^2\left(x+y\right)+\left(x-y\right)\left(x+y\right)+2x+2y+3\)

\(=-x^2+y^2+\left(-x+y\right)-2+3\)

\(=-\left(x-y\right)\left(x+y\right)-\left(x-y\right)+1\)

\(=\left(x-y\right)\left(-x-y-1\right)+1\)

\(=\left(x-y\right)\left(1-1\right)+1=1\)

`a)(x-1)(x^2+x+1)`

`=x^3+x^2+x-x^2-x-1`

`=x^3-1`

`b)(x^3+x^2y+xy^2+y^3)(x-y)`

`=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4`

`=x^4-y^4`

a) VT`=(x-1)(x^2+x+1)`

`=x^3 +x^2 +x -x^2-x-1 `

`=x^3-1=` VP.

b) VT `=(x^3+x^2y+xy^2+y^3)(x-y)`

`=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4`

`=x^4-y^4=` VP.

= (√x - √y)(√x + √y)2

= (√x - √y)(√x + √y)(√x + √y)

= (x - y)(√x + √y)

Đáp án: A

D   =   ( x 3   +   y 3 )   –   x y ( x   +   y )     =   ( x   +   y ) ( x 2   –   x y   +   y 2 )   –   x y ( x   +   y )     =   ( x   +   y ) ( x 2   –   x y   +   y 2   –   x y )     =   ( x   +   y ) [ x ( x   –   y )   –   y ( x   –   y ) ]     =   ( x   +   y ) ( x   –   y ) 2

Vì x = y ó x – y = 0 nên D   =   ( x   +   y ) ( x   –   y ) 2   =   0

Đáp án cần chọn là: D

a) ab + b√a + √a + 1 = [(√a)2b + b√a] + (√a + 1)

= b√a(√a + 1) + (√a + 1) = (√a + 1)(b√a + 1)

= (√x - √y)(√x + √y)2

= (√x - √y)(√x + √y)(√x + √y)

= (x - y)(√x + √y)